Problem: You are given an unweighted tree with n vertices. Recall that a tree is a connected undirected graph without cycles.

Your task is to choose three distinct vertices a, b, c on this tree such that the number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c is the maximum possible. See the notes section for a better understanding.

The simple path is the path that visits each vertex at most once.

Input

The first line contains one integer number n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree. 

Next n - 1 lines describe the edges of the tree in form a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). It is guaranteed that given graph is a tree.

Output

In the first line print one integer res — the maximum number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c.

In the second line print three integers a, b, c such that 1 ≤ a, b, c ≤ n and a ≠, b ≠ c, a ≠ c.

If there are several answers, you can print any.

Example

Input


8
1 2
2 3
3 4
4 5
4 6
3 7
3 8


Output


5
1 8 6

Note

The picture corresponding to the first example (and another one correct answer):

<image>

If you choose vertices 1, 5, 6 then the path between 1 and 5 consists of edges (1, 2), (2, 3), (3, 4), (4, 5), the path between 1 and 6 consists of edges (1, 2), (2, 3), (3, 4), (4, 6) and the path between 5 and 6 consists of edges (4, 5), (4, 6). The union of these paths is (1, 2), (2, 3), (3, 4), (4, 5), (4, 6) so the answer is 5. It can be shown that there is no better answer.

Solution (in PYTHON3):

Solution:
from collections import namedtuple, deque
from sys import stdin

def load_graph():
    n = int(stdin.readline())
    G = [set() for i in range(n)]
    for line in stdin.readlines():
        a, b = line.split()
        a, b = int(a) - 1, int(b) - 1
        # ~ print(a, b)
        G[a].add(b)
        G[b].add(a)
    return G


# find the distance from one node to any other node
# complexity O(n)
def find_all_distances(G, node):
    distances = [None] * len(G)
    distances[node] = 0
    queue = deque((node,))
    while len(queue) > 0:
        current_node = queue.popleft()
        for neighbor in G[current_node]:
            if distances[neighbor] is None:
                distances[neighbor] = distances[current_node] + 1
                queue.append(neighbor)
    # ~ print(distances)
    return distances


# find a pair of nodes with maximal distance
# G must be a tree, complexity O(n)
# strategy: pick node 0 as root, find a node with maximal distance to 0
# from that node, find a node with maximal distance to it
# it can be shown this pair of two nodes has maximal distance
def find_longest_pair(G):
    best_dist_0 = max((d, i) for i, d in
                       enumerate(find_all_distances(G, 0)))
    start_node = best_dist_0[1]
    best_dist_start = max((d, i) for i, d in
                          enumerate(find_all_distances(G, start_node)))
    path_length, end_node = best_dist_start
    return path_length, start_node, end_node


# find a triplet of distinct nodes
# with maximized sum of pairwise distances
# G must be a tree, complexity O(n)
def find_maxdist_triplet(G):
    path_length, path_a, path_b = find_longest_pair(G)
    dists_a = find_all_distances(G, path_a)
    dists_b = find_all_distances(G, path_b)
    # ~ print(path_a, dists_a, path_b, dists_b)
    sum_dists = [(dists_a[i] + dists_b[i], i)
                 for i in range(len(G)) if i != path_a and i != path_b]
    # ~ print(sum_dists)
    best_c_dist_ab, best_c = max(sum_dists)
    dist_sum = path_length + best_c_dist_ab
    return dist_sum, path_a, path_b, best_c


G = load_graph()
# ~ print(G)
dist_sum, path_a, path_b, path_c = find_maxdist_triplet(G)
print(dist_sum // 2)
print(path_a + 1, path_b + 1, path_c + 1)